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Does $\forall n\in \mathbb N: |A_n| \geq 2^n =|\{0,1\}^n|$ imply that $\underset {n\to \infty} \lim |A_i| \geq |\{0,1\}^\mathbb N|$

This is a lemma I would need to get around some confusing Nested Intervals in a proof of Cantor-Staub, but I am not sure if it is correct. (I know I should just learn nested intervals, but I'm told good mathematicians are also lazy so...)

I went over this again, and I realised, there is some context missing:

$A_0 := [0, 1]$

$A_1 := [0, 1/3] \cap[2/3, 1]$

$A_2 := [0, 1/9] \cap[2/9, 3/9] \cap [6/9, 7/9] \cap[8/9, 1]$

And so forth, each step cutting out a third of the largest continuous sub-intervals.

Now trying to show that $\bigcap_{i =0}^\infty A_i = |\{0,1\}^\mathbb N|$.

  • Now as $\mathbb R$ is compact we know that $|[a,b]| \geq 1$ for $a\leq b$ call it $(i)$
  • $\bigcap_{i=1}^n A_i = A_n$, as $\forall n \in \mathbb N:A_{n+1} \subset A_{n}$, so $\bigcap_{i=0}^\infty A_i = \underset {n\to \infty} \lim \bigcap_{i=0}^n A_i = \underset {n\to \infty} \lim A_n$
  • Per $(i)$ it holds that $A_n \geq 2^n =|\{0,1\}^n|$, the number of biggest continuous intervals in $A_i$ doubles in every step ($i\to i+1$) .
  • Now the question is, does this "invariant" persist in the limit, such that $\underset {n\to \infty} \lim A_n \geq \underset {n\to \infty} \lim|\{0,1\}^n|$?

Intuitively I would say yes, but this is very dangerous territory, as $\underset {n\to \infty} \lim 2^n$ is not properly defined...

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    $\begingroup$ What if for some $k$ for all $n\geq k$ we have $|A_n|=|\mathbb N|$? $\endgroup$
    – amrsa
    May 15 at 14:19
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    $\begingroup$ For example, if $A_n=\mathbb N$, then $|A_n|\geq 2^n$, but $\lim|A_n|=|\mathbb N|<|\{0,1\}^{\mathbb N}|$. $\endgroup$
    – amrsa
    May 15 at 14:21
  • $\begingroup$ Thanks @amrsa You just answered my question. Thanks. (I feel a little ashamed that I didn't figure it out myself...). Is there some way to equate the growth of a set to the growth of a binary string? Say if $|A_n| = \Omega(2^n)$, i.e. $|A_n|$ grows at least as fast as $2^n$. I can see how this is not helpful notation, as this kind of calculus with infinity doesn't really work... $\endgroup$ May 15 at 14:51
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    $\begingroup$ Your set theory notation looks off to me. You write for example $$A_0 := \{[0,1]\}$$ This means that $A_0$ is a set of cardinality $1$, having a unique element (namely the set $[0,1]$). Similarly $A_1$ is a set of cardinality $1$, having a unique element (namely the set $[0,1/3] \cup [2/3,1]$). I suspect this is not what you intended, I suspect instead that you intended $A_0 = [0,1]$, $A_1 = [0,1/3] \cup [2/3,1]$, and so on. As currently written, $A_0$ and $A_1$ have no object in common, so $A_0 \cap A_1 = \emptyset$, hence $\bigcap_{i=0}^\infty A_i = \emptyset$. $\endgroup$
    – Lee Mosher
    May 15 at 14:51
  • $\begingroup$ Thanks for noticing @LeeMosher, you are of course right. I adjusted it $\endgroup$ May 15 at 14:54

1 Answer 1

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Thanks to @amrsa's comment I have the answer before I could even type out my question properly ^^ For $\forall n:A_n = \mathbb N$ we have $|A_n| \geq 2^n$ but $\lim |A_n| = |\mathbb N| < |\{0,1\}^{\mathbb N}|$

So, the answer is no.

I wonder why my intuition was so wrong about it.

One needs to be careful with infinities I suppose?

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